$B_s$ is brownian motion. Because $\int_{t-1}^t e^{-μ(t-s)}\sigma dB_s $ has a Brownian component, it is normally distributed according to Taylor & Karin 1998 's Introduction to Stochastic Modelling.
The above equation is derived from the ornstein uhlenbeck model: $$dx_{t}= μ(θ-x_t)dt+\sigma B_t $$
and the above equation is $x_t$ given $x_{t-1}$.
Based on Leung & Li 2016's Optimal Mean Reversion Trading, the distribution of $x_t$ given $x_{t-1}$ under the Ornstein Uhlenbeck model is normally distributed with $$E[x_t|x_{t-1}]=x_{t-1}e^{-μΔt}+ θ(1-e^{-μΔt})$$
$$ Var(x_t|x_{t-1})= \frac{\sigma^2}{2μ}(1-e^{-2μΔt})$$
How can I prove its normally distrbuted? I already derived the mean and variance in my own time, but I still dont know how to prove its normally distributed.
I saw this question with its answer:
Ornstein-Uhlenbeck stochastic process and gaussian process
"Depending on the precise context, we might be able to use the following theorem:
If :[0,]→ℝ
is nonrandom and is a Gaussian process for 0≤≤, then ()+ is a Gaussian process for 0≤≤"
Can this be a clue? Please help, thanks!